Answer
$$\frac{1}{3} \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2}$$
Work Step by Step
Since$$ f^{\prime}(x)=\frac{15 * x^{2}}{\sqrt{5 x^{3}+4}} \geq 0$$
then $f(x)$ is increasing on $ [ 0,1] $ and $ \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}}$ has an upper bound at $x=0$ $$\int_{0}^{1} \frac{1}{\sqrt{5(0)+4}}=\int_{0}^{1} \frac{1}{2} d x=1 / 2$$
and a lower bound
$$\int_{0}^{1} \frac{1}{\sqrt{5(1)+4}}=\int_{0}^{1} \frac{1}{3} d x=1 / 3$$
Then, by using the comparison theorem
\begin{aligned}
\int_{0}^{1} \frac{1}{3} d x & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \int_{0}^{1} \frac{1}{2} d x \\
\frac{1}{3} *(1-0) & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2} *(1-0) \\
\frac{1}{3} & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2}
\end{aligned}
